im so mad, how do i solve this?

You frist factor it and then set the factor equal to zero to obtain the "critical points" this is when the graph goes from being less then zero to more than zero, and other way around.

(x-8)(x-4) < 0

x=8 and x=4 critical points.....

now test point between the critical points and see if they are less then of more than zero. Once you tests points from each section between the points you'll find out x is less than zero between 4 and 8, or 4<x<8 or (4, 8)

If this equation was = 0, it would be factored as:

(x - 8)(x - 4) or x = 8; x = 4. Since the sign is less than (<) 0 then the equation doesn't work out correctly.

If you factorize the given quadratic expression you get (x-8)(x-4)

=> (x-8)(x-4) < 0

Taking 4 and 8 as critical points,

For x < 4....the L.H.S is positive whereas the R.H.S says < 0

So x cannot be less than 4.

For x > 8...the L.H.S is again positive whereas the R.H.S says < 0

So x cannot be greater than 8 also.

Therefore must lie between 4 and 8,i.e. 4 < x < 8.

You can put any value for x lying between 4 and 8 (say,3) and check that the above solution is correct.

x^2 -12x + 32 < 0

(x - 8)(x-4)<0

If multiplying two things gives us a negative number, then 1 (but not both) of the factors is less than 0.

so 4<x<8

You messed it up...it is one isn't a dime and it is 3 money. So it is going like this....a guy has 3 money that make up 25 cents, one isn't a dime..what are they? 2 Dimes and a million Nickle

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it would factor to (x-4)(x-8)<0

x=4 and x=8

so i guess you make an interval

that x exists in the interval of (4,8)

O_O